Refrigerator Binomial Distribution problem

Question

A company produces refrigerators and knows that the proportion of defective units is 0.05.

The refrigerators are sold in lots of 35 units. A large network of appliance stores accept the lot if it has a maximum of 2 refrigerators with defect in the batch.

Admitting that they've accepted the lot, what is the probability of having observed exactly one faulty refrigerator?

Answer: 0.4102

I've tried to do this using the binomial distribution "formula" using N (trials) as 35 and n (success) as 1, and 2, and probability of success as 0.05 and couldn't resolve this question.

Answer 1

The binomial distribution's pmf is: $p(x)=\binom{35}{x}(0.05)^x(0.95)^{35-x} \quad[0\leq x\leq 35, x\in\Bbb N]$

The conditional probability that there is exactly 1 faulty unit when given that there is a maximum of 2 is $$\dfrac{p(1)}{p(0)+p(1)+p(2)}$$

Answer 2

Given that the store has accepted the batch, there must be 0, 1, or 2 failures. Using the binomial distribtuon function, these have a priori chances of 0.1661, .3059, and 0.2737 respectively. The chances of exactly one is therefore 0.3059/(0.1661 + 0.3059 +0.2737) = 0.4102.